How do you find the product #(4x+5)(4x+5)#?

1 Answer
Jun 15, 2018

Answer:

See a solution process below:

Explanation:

This problem can be rewritten as:

What is: #(4x + 5)^2#

This is a special form of the quadratic which can be expanded using the rule:

#(color(red)(x) + color(blue)(y))^2 = (color(red)(x) + color(blue)(y))(color(red)(x) + color(blue)(y)) = color(red)(x)^2 + 2color(red)(x)color(blue)(y) + color(blue)(y)^2#

Substitute #color(red)(4x)# for #color(red)(x)# and #color(blue)(5)# for #color(blue)(y)# gives:

#(color(red)(4) + color(blue)(5))^2 = (color(red)(4) + color(blue)(5))(color(red)(4) + color(blue)(5)) = (color(red)(4x))^2 + (2 * color(red)(4x) * color(blue)(5)) + color(blue)(5)^2 = 16x^2+ 40x + 25#